Turns out this integral has a very nice closed form:

$$\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$$

I found it with Mathematica, but I can't figure out how to prove it.

The integral seems quite problematic to me. If the limits were finite, I would do this:


Then, for one of the integrals we will have:

$$\int_A^B \sqrt{x^4+a^4} dx=a^3 \int_{A/a}^{B/a} \sqrt{1+t^4} dt$$

This integral is complicated, but quite well known.

On the other hand $\int_0^\infty \sqrt{1+t^4}dt$ diverges, so I can't consider the two terms separately.

But the integral behaves like I can! If we look at the final expression, it seems like $\int_0^\infty \sqrt{1+t^4}dt=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}}$ even though it can't be correct.

I have to somehow arrive at Beta function, since we have a squared Gamma as an answer.

I'm interested in this integral, since it represents another kind of mean for two numbers $a$ and $b$. If we scale it appropriately:

$$I(a,b)=\frac{8 \sqrt{\pi}}{\Gamma(1/4)^2 } \int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}= \frac{4}{3} \frac{a^2+ab+b^2}{a^3+ab(a+b)+b^3}$$

So, $1/I(a,b)$ is a mean for the two numbers.

  • for your question after the separation line, I totally have no ideas but I hold such a doctrine that it may be a coincidence. – Zack Ni Aug 21 '16 at 3:21
  • 1
    @ZackNi, there is no question after the separation line. I just made a statement – Yuriy S Aug 21 '16 at 12:53
  • Okay I misunderstood it. – Zack Ni Aug 21 '16 at 13:00
up vote 5 down vote accepted

$$\int_0^\infty (\sqrt{x^4+a^4}-\sqrt{x^4+b^4}) dx \implies $$

$$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx$$

Because the $\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx $ is convergent so the integration can be linearly seperated.

$$\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx - \int_0^\infty(\sqrt{x^4+b^4}-x^2) dx$$

Out of symmetry, the question becomes how to solve integration $\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx$ in terms of $a$.

Let $a^2 \sqrt{t} = \sqrt{x^4+a^4}-x^2 \implies x^2 = a^2 \frac{1-t}{2\sqrt{t}}$ which brought us to an old problem: see the second answer in this post by votes sort.

After simplification the integration becomes $\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} a^3$.

Therefore, the original formula should be $$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx = \frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} (a^3-b^3)$$.

Hence the question get solved.

  • Nice! I was looking for a complex analytic approach but couldn't find anything. – iamvegan Aug 21 '16 at 2:56
  • @iamvegan Thanks. Because this method is very suitable for human knowing the math: from specific simple example to the complex one and I think WA is really a magnate of calculus fields, such simple transformation it can achieve. – Zack Ni Aug 21 '16 at 3:01
  • Thank you. Although I'm not sure how are we allowed to add and subtract a divergent integral $\int_0^\infty x^2 dx$ – Yuriy S Aug 21 '16 at 12:54
  • @yuriys subtract is allowed under integral operation that's why I don't combine the second step and third step but split the integral is not allowed. consider this example $\int_0^\infty 0 dx = \int_0^\infty x^2-x^2 dx = 0$ But $\int_0^\infty x^2 dx- \int_0^\infty x^2 dx $ is not allowed. – Zack Ni Aug 21 '16 at 12:59

Another way to do it, may be.

Assuming $a>0$ and $t>0$, we have

$$I_a=\int_0^t \sqrt{x^4+a^4}\, dx=a^2 t \, _2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{t^4}{a^4}\right)$$ Expanding as Taylor series for infinitely large values of $t$, $$I_a=\frac{t^3}{3}+\frac{\sqrt{\frac{\pi }{2}} a^3 \Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{7}{4}\right)}-\frac{a^4}{2 t}+O\left(\frac{1}{t^3}\right)$$ So, for an infinite value of $t$,
$$\lim_{t\rightarrow\infty}(I_a-I_b)=(a^3-b^3)\frac{\sqrt{\frac{\pi }{2}} \Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{7}{4}\right)}=(a^3-b^3)\frac{\Gamma \left(\frac{1}{4}\right)^2}{6 \sqrt{\pi }}$$ and hence the result.

Another way to split the integrals is to replace the square root by a power $p$:

$$\sqrt{x^4 + a^4}\longrightarrow \left(x^4 + a^4\right)^{p}$$

The integral of the separate terms will then converge for $p<-\frac{1}{4}$ and can be expressed in terms of the beta-function. You can substitute $p = \frac{1}{2}$ in the final answer, despite the individual integrals not converging by invoking analytic continuation.

To get to the beta-functions, you can substitute $x = a t$ to get $a$ out of the way, then $u = t^4 + 1$ and finally $u = \frac{1}{v}$ will yield the explicit beta-function form.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.